﻿ Beyond forces: work, kinetic energy, and potential energy | Bits of Pancake

Beyond forces: work, kinetic energy, and potential energy

I learned about stuff like the work and kinetic energy in high school, but I never really understood why theses concepts were introduced — it feels like they just fell out of the sky. That changed today in physics lecture, and I think it’s so cool to understand where exactly these concepts come from.

So let’s say we’re working with forces in physics. If you look at some common forces, like gravity and the electromagnetic force, we see that forces tend to depend on position:

$$\begin{gather} \vecb{F}(\vecb{r}) = \frac{GMm}{r^2} \hatb{r} \\ \vecb{F}(\vecb{r}) = \frac{kq_1q_2}{r^2} \hatb{r} \end{gather}$$

But force is defined using a time derivative of velocity:
$$\vecb{F}(t) = m\vecb{a} = m\dod{\mathbf{v}}{t}$$

This is kind of inconvenient. The definition of force depends on time, but most forces are expressed with respect to position. Meanwhile, the positions of particles tend to vary with time (i.e., they move), making things more complicated. In order to work with moving particles experiencing forces, we would need to convert all of our times into positions or vice-versa.

It turns out that there’s an easier way. Someone clever decided to integrate both sides of Newton’s second law along the position to get these line integrals along a path $C$:
\begin{align} \int_C \vecb{F} \cdot \dif{\vecb{r}} &= m \int_C \dod{\vecb{v}}{t} \cdot \dif{\vecb{r}} \\ &= m \int_{t_1}^{t_2} \dod{\vecb{v}}{t} \cdot \vecb{v}\dif{t} \\ &= m \int_C \vecb{v} \cdot \dif{\vecb{v}} \\ &= m \left( \int_{v_{x1}}^{v_{x2}} v_x \dif{v_x} + \int_{v_{y1}}^{v_{y2}} v_y \dif{v_y} + \int_{v_{z1}}^{v_{z2}} v_z \dif{v_z} \right) \\ &= \frac{1}{2}m \left( \left(v_{x2}^2 – v_{x1}^2\right) + \left(v_{y2}^2 – v_{y1}^2\right) + \left(v_{z2}^2 – v_{z1}^2\right) \right) \\ &= \frac{1}{2}mv_2^2 – \frac{1}{2}mv_1^2 \\ \end{align}

And we’re left with the Work-Energy Theorem:
$$\int_C \vecb{F} \cdot \dif{\vecb{r}} = \frac{1}{2}mv_2^2 – \frac{1}{2}mv_1^2$$

Now, if we define the quantity $\frac{1}{2}mv^2$ to be the “kinetic energy” and the quantity $\int \vecb{F} \cdot \dif{\vecb{r}}$ to be the “work” (the “total” force applied to a system over a distance), then we can express the theorem intuitively: work applied to a system increases the kinetic energy. And now we can use it to solve problems without having to deal with times — only distances.

Let’s try a quick example. If gravity applies a force of 10 N to a box from rest for 10 m, what will its velocity be? In this case, $F = 10 \text{ N}$, $r_2 = 10 \text{ m}$, and $r_1 = 0 \text{ m}$. We can now plug in (we can replace the line integral with a regular integral, assuming we’re moving in a straight line):
$$\begin{gather} \int_{r_1}^{r_2} F \dif{r} = \frac{1}{2}mv_2^2 – \frac{1}{2}mv_1^2 \\ F(r_2 – r_1) = \frac{1}{2}mv_2^2 – \frac{1}{2}mv_1^2 \\ \end{gather}$$
And from there, we can solve for velocity. Again, notice that we don’t care about the time it takes, just the distance over which a force acts. Cool! (What if we cared about the time it takes rather than the distance? Then we can use impulse, which comes from the definition of force.)

There’s one problem with the Work-Energy Theorem: evaluating the line integral is annoying. Converting that line integral into a regular integral is fine for the example I just gave, but it won’t work if the path is not in the same direction as the force.

It turns out, though, that a lot of the forces we care about (gravity, springs, electromagnetic force) are conservative forces. This means that the line integral is path-independent: its value doesn’t depend on which path you take from point A to point B; it only matters where point A and point B are.

One analogy is with regular integrals. If you’re integrating from $x = 0$ to $x = 5$, you’ll always get the same answer as long as the ultimate endpoints are the same.
$$\int_0^5 f(x) \dif{x} = \int_0^{-10} f(x) \dif{x} + \int_{-10}^7 f(x) \dif{x} + \int_7^5 f(x) \dif{x}$$
Along the same lines, for line integrals of conservative forces, you’ll always get the same answer no matter which path you take.

The path-independence of regular integrals is what makes the fundamental theorem of calculus possible:
$$\int_a^b f(x) \dif{x} = F(b) – F(a)$$

It turns out that there’s an analogous version of the fundamental theorem for line integrals of conservative forces:
$$\int_C \vecb{F}(\vecb{r}) \cdot \dif{\vecb{r}} = U(\vecb{r}_2) – U(\vecb{r}_1)$$ where $U$ is the “antiderivative” — the scalar potential of the vector field $\vecb{F}$. In other words, for every conservative force, we can define a potential function that we can use to evaluate line integrals of conservative forces. This potential function will have a value for every point in space. (Note that the value will be a scalar and not a vector.) For example, for the force of gravity, $\vecb{F}(\vecb{r}) = \frac{GMm}{r^2}\hatb{r}$, and $U(\vecb{r}) = -\frac{GMm}{r} + C$, where $C$ is the constant of integration, here usually taken so that $U(\infty) = 0$. Notice that $\dod{U}{r} = F$ (or the multidimensional equivalent, $\nabla{U} = \vecb{F})$!

So for a conservative force, the Work-Energy Theorem becomes:
$$U(\vecb{r}_2) – U(\vecb{r}_1) = \frac{1}{2}mv_2^2 – \frac{1}{2}mv_1^2$$ or
$$\frac{1}{2}mv_1^2 + U(\vecb{r}_1) = \frac{1}{2}mv_2^2 + U(\vecb{r}_2)$$
Let’s call $U$ the potential energy, and now we’ve obtained an expression for the conservation of mechanical energy. As long as we’re talking about conservative forces like gravity, the sum of the kinetic and potential energy stay the same no matter what you do.

So that’s how you get work, kinetic energy, and potential energy from forces. I hope they seem more connected now and less like a jumble of concepts that physics teachers throw at you! Leave a comment if it was helpful!

February 26, 2013, 5:01pm by Casey